giải phương trình sau
\(\left(\dfrac{x-a}{bc}-\dfrac{1}{b}\right)+\left(\dfrac{x-b}{ca}-\dfrac{1}{c}\right)+\left(\dfrac{x-c}{ab}-\dfrac{1}{a}\right)=\dfrac{ab+bc+ca}{abc}\)
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Rút gọn biểu thức:
a) \(A=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ca}{\left(b-c\right)\left(b-a\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
b) \(B=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+\dfrac{1}{x^3}}\)
Rút gọn biểu thức: \(B=\left(ab+bc+ca\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)
\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=2\left(a+b+c\right)\)
Cho a,b,c là các số thực dương thỏa mãn a+b+c=3. Tìm giá trị nhỏ nhất của biểu thức:
\(P=\dfrac{1}{a\left(b^2+bc+c^2\right)}+\dfrac{1}{b\left(c^2+ca+a^2\right)}+\dfrac{1}{c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}\)
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a,b,c là các số thực dương thỏa mãn a+b+c=1. Chứng minhh rằng:
\(\left(1+ab+bc+ca\right)\left(\dfrac{1}{a+bc}+\dfrac{1}{b+ca}+\dfrac{1}{c+ab}\right)\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
Áp dụng BĐT quen thuộc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)
Ta chỉ cần chứng minh:
\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)
Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)
Nên ta chỉ cần chứng minh:
\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)
Cho a,b,c>0 thỏa mãn : \(ab+bc+ca=0\)
C/m: \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge3+\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\dfrac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\dfrac{\left(c+a\right)\left(c+b\right)}{c^2}}\)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
may cai nay tuong hoi truoc co nguoi dang roi ma
ta có:
\(\sqrt{\dfrac{\left(a+b\right).\left(a+c\right)}{a^2}}\le\dfrac{1}{2}.\left(\dfrac{a+b}{a}+\dfrac{a+c}{a}\right)=a+\dfrac{b}{2}+\dfrac{c}{2}\)
tương tự thì ta có:
\(VP\le3+2\left(a+b+c\right)\)
\(VP=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=3+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}\)
từ các điều trên ta thấy cần CM:
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge a+b+c\)
bạn tự CM nốt ạ
Bài 1 : Tìm a,b,c biết :
a) Cho \(\dfrac{\overline{ab}+\overline{bc}}{a+b}=\dfrac{\overline{bc}+\overline{ca}}{b+c}=\dfrac{\overline{ca}+\overline{ab}}{c+a}\left(a,b,c\ne0\right)\). Tính \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
b) Cho a,b,c là các số thực khác 0 sao cho : \(\dfrac{2x+2y-z}{z}=\dfrac{2x-y+2z}{y}=\dfrac{x+2y+2z}{x}\). Tính giá trị của biểu thức \(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8.x.y.z}\)
cho a,b,c>0;\(a+b+c,abc=1\).CMR
\(\dfrac{bc}{a^2\left(b+c\right)}+\dfrac{ca}{b^2\left(c+a\right)}+\dfrac{ab}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
Chứng minh
\(\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(A=\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ca}\)
\(B=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
..................................
\(A=\dfrac{a^2+b^2-2ab}{ab}+\dfrac{b^2-2ab+c^2}{bc}+c^2+a^2-\dfrac{2ca}{ca}\)
\(A=\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}-2\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}-2\right)=\dfrac{\left(b+c\right)}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}-6\)
\(A=\left[\dfrac{\left(b+c\right)}{a}+1\right]+\left[\dfrac{\left(a+c\right)}{b}+1\right]+\left[\dfrac{\left(a+b\right)}{c}+1\right]-9\)
\(A=\dfrac{\left(a+b+c\right)}{a}+\dfrac{\left(a+b+c\right)}{b}+\left[\dfrac{\left(a+b+c\right)}{c}\right]-9\)
\(A=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-9\)
Ket luan
\(A\ne B\) => đề sai--> hoặc mình công trừ sai